In what follows $B$ denotes two-dimensional Brownian motion and $\mathbb{B}(0,z)$ denotes the disk with radius $z$ centered at the origin. I also define the hitting time $\tau_C = \inf\{t \geq 0: B_t \in C\}$. It then follows that
$$P^x\{\tau_{\mathbb{B}(0,r)} < \tau_{\mathbb{B}(0,R)^{\mathsf{c}}}\} = \frac{\log{R} - \log{\lvert x\rvert}}{\log{R} - \log{r}}$$
for $0 < r < \lvert x\rvert < R$. In other words, the expression above is the probability that the two-dimensional Brownian motion started at $x$ enters the ball with radius $r$ before it exits the ball with radius $R$.
Using this expression I want to compute $P^x\{\tau_{\{0\}} < \infty\}$, namely the probability that the two-dimensional Brownian motion started at $x$ hits the origin in finite time.
The double limit $$\lim_{R \to \infty}\lim_{r\to 0}\frac{\log{R} - \log{\lvert x\rvert}}{\log{R} - \log{r}} = 0$$ gives the right answer. But the double limit $$\lim_{r\to 0}\lim_{R \to \infty}\frac{\log{R} - \log{\lvert x\rvert}}{\log{R} - \log{r}} = 1$$ gives the wrong answer. I am trying to understand what is going wrong in this latter approach.
First, $$\lim_{R \to \infty}P^x\{\tau_{\mathbb{B}(0,r)} < \tau_{\mathbb{B}(0,R)^{\mathsf{c}}}\} = P^x\{\tau_{\mathbb{B}(0,r)} < \infty\} = 1$$
But somehow it is not true that $$1_{\tau_{\mathbb{B}(0,r)}(\omega) < \infty}(\omega) \to 1_{\tau_{\{0\}}(\omega) < \infty}(\omega) \label{a}\tag{1}$$ as $r \to 0$. If it were true, then $P^x\{\tau_{\{0\}} < \infty\}$ would be $1$, which is wrong. But I don't see why $\ref{a}$ does not hold. Can someone explain the error in my reasoning?